graphing rational functions

[jinx24]

I am really having trouble with this one. I know it is a fairly easy one and I am probably making it harder that what it really is.

"Sketch the graph of the rational function and specify the intercepts and the asymptotes"

y = -3/(x + 2)

So far I have:
y-int = -3/2
HA = y=0
VA = x=-2

I am having trouble with the x-intercept. Well, I know the x-intercept is -3, what is counfusing me though, is that when I checked my answer on the graphing calculator, the graph did not pass through the x-intercept. How do I know when there is an x intercept and when there is not?

Should I be taking a different approach to this problem...like using tansformations? If so, how would I handle the numerator?

Thank you

 

[galactus]

Set your equation to y=0 and solve for x to find the x-intercept. In this

case, there is no x-intercept. The graph approaches the x axis but never

intercepts it.

Do you see why?.

 

[jinx24]

No, I don't see why the graph never intercepets the x axis. I see that the x axis is the horizontal asymptote. Maybe I am soving for x wrong?

0 = -3/(x + 2)
(x + 2) = -3
x = -5

Thank you

 

[galactus]

Yes, you are solving for x wrong. You can not multiply x+2 times 0 and

get x+2. For that matter, check to see if your -5 is correct.

\(\displaystyle \frac{-3}{(-5)+2}=1\), not 0.

The reason there is no x-intercept is because no matter what value you

enter into x, y will get closer to 0 but never quite make it.

Try it:

\(\displaystyle \frac{-3}{100+2}=-.029\)

\(\displaystyle \frac{-3}{100000000+2}=-.000000029999.........\)

etc, etc.

 

[jinx24]

Thank you so much. I get it now. I have been doing so many problems where I have to find where the graph crosses the HA, which is usually 1...so in this problem I forgot to multipy by 0. Thank you, again, for pointing that out to me

Jenny